[webassignbuddy]

I did: Kp = Kc x (RT)^(n_g)

KP = (0.25/3^4)(0.08206 x 1325)^-3
Kp = (0.00308642)(108.7295)^-3
Kp = 2.0 x 10^-19 (INCORRECT)

this is frustrating bc the answer is supposed to be 324....

[sjb]

Have you used the right R?

[Borek]

I did: Kp = Kc x (RT)^(n_g)

And where did you got Kc from?

KP = (0.25/3^4)(0.08206 x 1325)^-3

0.25 and 3.0 are pressures, not concentrations.

[webassignbuddy]

Have you used the right R?

Yes that's what my textbook used in a similar example!

[webassignbuddy]

I did: Kp = Kc x (RT)^(n_g)

And where did you got Kc from?

KP = (0.25/3^4)(0.08206 x 1325)^-3

0.25 and 3.0 are pressures, not concentrations.

I know but you do

Kp = [P_S8]/[P_S2]⁴
Kp = [S₈](RT)/[S₂]⁴(RT)⁴
Kp = ([S₈]/[S₂⁴])(RT)^3-

Edit:

Oh wait!!! I was looking at the wrong answer key...
the answer is 3.08 x 10^3-

So all I do is

Kp = [P_S8]/[P_S2]⁴
Kp = 0.25/3⁴
Kp = 0.003086
Kp = 3.08 x 10^3-

!!!

[webassignbuddy]

I did: Kp = Kc x (RT)^(n_g)

And where did you got Kc from?

KP = (0.25/3^4)(0.08206 x 1325)^-3

0.25 and 3.0 are pressures, not concentrations.

For part B, I found:
Q = [S₂]/[S₈]⁴
Q = (1.8 x 10^-3)/(9.76 x 10^-1)⁴
Q = 1.98 x 10^-3 (CORRECT)

Since Q < K, ΔG is negative and the reaction is SPONTANEOUS (to the right) (CORRECT)

How do I calculate Kc though? Because it's not Kc = [S₂]/[S₈]⁴

[Borek]

You know how to find Kp when Kc is given, but you don't know how to find Kc when Kp is given?

[webassignbuddy]

You know how to find Kp when Kc is given, but you don't know how to find Kc when Kp is given?

Ohh!!

Kc = Kp x (RT)^-Δn_g
Kc = 3.96 x 10³

Thanks!

I may have one more question!

[webassignbuddy]

You know how to find Kp when Kc is given, but you don't know how to find Kc when Kp is given?

How would I go about finding ΔG°?

[webassignbuddy]

You know how to find Kp when Kc is given, but you don't know how to find Kc when Kp is given?

do you know how I can find ΔG?

[webassignbuddy]

Anyone?